Let $n=2a$ for some $a\in\mathbf{N}_{>1}$ so that $4a^2=p+qr$. If $p=2$, then either $q$ or $r$ must equal $2$. Suppose it is $q$, then $2a^2=1+r\Leftrightarrow r=2a^2-1=2a^2-2+1=2(a+1)(a-1)+1$. Clearly, $(a+1)(a-1)$ is composite for $a>2$.
I noticed that the equation in particular, $r=2a^2-1$, obeys a negative pell equation
I can prove that if $n=2a$ for some natural number $a>1$ (greater than $1$ because $n>2$), then there are certain values that $a$ can never equal if $p=q=2$.